（树型dp，数论）多校7 Mahjong tree

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5379

官方题解：

##1011. Mahjong tree

在一棵树上给所有点标号，要求任意一个子树里的点编号连续，每一个点的儿子编号连续。那么，一个点的非叶子儿子应该是连续的，即一个点的非叶子儿子最多只有两个。对于每一个点，我们把它的叶子儿子的个数记作S，所有儿子的方案数积为T。当非叶子儿子节点个数小于2的时候，方案数为2T*(S!). 当非叶子儿子节点数等于2的时候，这个点为根的子树合法方案数位T*（S!）. 这样dfs一遍即可以处理整棵树的方案数。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <queue>#include <vector>#include <set>#include <algorithm>using namespace std;#define mem(x,i)memset(x,i,sizeof(x))#define sfi(a)scanf("%d", &a)  #define sfii(a,b)scanf("%d %d", &a, &b)  #define sfiii(a,b,c)scanf("%d %d %d", &a, &b, &c)  #define sfscanf#define pfprintf const double EPS = 1e-10;const double pai = acos(-1.0);const int INF = 0xfffffff;const int MOD = 1000000007;typedef long long LL;const int maxn = 1005;int N, K;vector<int> g[maxn];int sz[maxn];LL dp[maxn][maxn], F[maxn], inv[maxn];void init(){for (int i = 1; i <= N; i++){g[i].clear();sz[i] = 0;}}LL pow_mod(LL a, LL p, LL n){if (p == 0) return 1;LL ans = pow_mod(a, p / 2, n);ans = ans*ans%n;if (p % 2 == 1) ans = ans*a%n;return ans;}void MakeInv(){for (int i = 1; i < maxn; i++){inv[i] = pow_mod(i, MOD - 2, MOD);}}void MakeF(){F[0] = 1;for (int i = 1; i < maxn; i++){F[i] = (F[i - 1] * i) % MOD;}}int dfs(int u, int fa){sz[u] = 1;for (int i = 0; i < g[u].size(); i++){int v = g[u][i];if (v == fa)continue;sz[u] += dfs(v, u);}return sz[u];}void solve(){dfs(1, -1);mem(dp, 0);dp[0][0] = 1;for (int i = 1; i <= N; i++){for (int j = 0; j <= K; j++){dp[i][j] += (dp[i - 1][j] * (sz[i] - 1) % MOD)*inv[sz[i]] % MOD;if (j > 0)dp[i][j] += dp[i - 1][j - 1] * inv[sz[i]] % MOD;dp[i][j] %= MOD;}}}int main(){//freopen("f:\\input.txt", "r", stdin);MakeInv();MakeF();int T;sfi(T);for (int cas = 1; cas <= T; cas++){sfii(N, K);init();for (int i = 0; i < N - 1; i++){int a, b;sfii(a, b);g[a].push_back(b);g[b].push_back(a);}solve();LL ans = dp[N][K] * F[N] % MOD;pf("Case #%d: %lld\n", cas, ans);}return 0;}

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